# Why is the pressure correction in the van der Waals equation proportional to (n/V)^2?

The van der Waals equation for real gases is stated as follows:

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

For the coefficient $b$, we can reason out that more the number of molecules, the more volume will be occupied by the molecules (in turn reducing the free space available for the motion of the molecules). It must be linearly proportional to $n$ because $b$ represents the sum of the excluded volume of each molecule in a mole. Therefore, we must multiply it by $n$ to obtain the total excluded volume.

The coefficient $a$ is said to represent the strength of the intermolecular attractive forces. Intuitively, we can say that it must be proportional to the number of molecules per unit volume. The more the number of molecules around a molecule, the more attractive forces it shall feel.

Therefore, the correction term for pressure must be proportional to $\frac{n}{V}$. But we know that in reality, it is proportional to the square of $\frac{n}{V}$.

Why does it vary squarely?

• – Zhe Mar 16 '17 at 14:11
• However the pressure term is modified and explained, it is worth remembering that the parameter a is intuitive rather than exact and that the van der Waals equation does not have sound theoretical basis. – porphyrin Mar 19 '17 at 20:51

Summing up the argument from the Wikipedia derivation, the reason it is proportional to $$(\frac{1}{V_\mathrm m})^2$$ where $$V_\mathrm m=\frac Vn$$ (the molar volume) comes from how the molecules experience the attractive force. The only molecules that experience a net attractive force are those close to the edge of the container because as we move toward the center of the container, the attractive forces on either side of a molecule cancel out. The force toward the center experienced by a molecule at the edge should be proportional to $$(\frac{1}{V_\mathrm m})$$. But we can also say that the number of molecules at the edge of the container should be proportional to $$(\frac{1}{V_\mathrm m})$$. So, by combining the proportionality constants into a single constant $$a$$, we obtain that the correction to the pressure should be $$(\frac{a}{V_\mathrm m^2})=\frac {an^2}{V^2}$$ .

Pressure correction term depends upon:

1. Number of molecules attracting the molecules which comes to strike the wall and as such it is proportional to density of gas i.e. proportional to $$n/V$$ where $$n$$ is the number of moles of gas and $$V$$ is the volume of the container.

2. It also depends upon number of molecules which has a strike the unit area of the wall and is therefore proportional to the total number of molecules per unit volume, i.e. proportional to the density again (i.e. proportional to $$n/V$$).

So the pressure correction term is jointly proportional to $$n/V$$ for factor 1 and again proportional to $$n/V$$ for factor 2. Hence the pressure correction term is proportional to

$$\left(\frac{n}{V}\right)\left(\frac{n}{V}\right),$$

i.e. proportional to

$$\frac{n^2}{V^2}.$$

As Tyberius wrote, the pressure reduction is due to a force imbalance that exists at the walls of the container, because a molecule hitting the wall doesn't feel attraction to the wall, but is attracted to the molecules on its other sides. The attractive forces from these molecules "pull back" on the molecule hitting the wall, reducing the pressure.

The reduction in the force with which each molecule hits the wall is proportional to the strength of intermolecular attractive force times the concentration of molecules pulling back on it, i.e., to $$\rho=n/V$$. This gives the force reduction per molecule.

The total pressure reduction is the force reduction per molecule ($$\sim \rho=n/V$$) times the number of molecules per unit area hitting the wall (also $$\sim \rho=n/V$$). Hence the pressure reduction due to intermolecular attraction is $$\sim \rho^2=(n/V)^2$$.